Question

a game at a carnival has the following rules: a player rolls a 6-sided die and, if the number is less than 6, wins twice the number on the die in tokens. on the other hand, the player loses 12 tokens when a 6 is rolled. what should it cost to play if the game is to be fair? 7 tokens for one game 9 tokens for two games 3 tokens for one game 1 token for two games

Explanation:

Step1: Calculate expected value

First, find the probability of each outcome. A 6 - sided die has 6 possible outcomes. The numbers less than 6 are 1, 2, 3, 4, 5 (5 outcomes), and 6 is 1 outcome. So the probability of rolling a number less than 6, \(P(X < 6)=\frac{5}{6}\), and the probability of rolling a 6, \(P(X = 6)=\frac{1}{6}\).

For \(X = 1\): win \(2\times1 = 2\) tokens.
For \(X = 2\): win \(2\times2 = 4\) tokens.
For \(X = 3\): win \(2\times3 = 6\) tokens.
For \(X = 4\): win \(2\times4 = 8\) tokens.
For \(X = 5\): win \(2\times5 = 10\) tokens.
For \(X = 6\): lose 12 tokens (so the value is - 12).

The expected value \(E(X)\) is calculated as:

\[

$$\begin{align*} E(X)&=\frac{1}{6}(2 + 4+6 + 8+10)+\frac{1}{6}(- 12)\\ &=\frac{1}{6}(30)+\frac{1}{6}(-12)\\ &=\frac{30 - 12}{6}\\ &=\frac{18}{6}\\ &= 3 \end{align*}$$

\]

Wait, no, wait. Wait, the probability for each of 1 - 5 is \(\frac{1}{6}\) each. So actually:

\[

$$\begin{align*} E(X)&=\frac{1}{6}(2)+\frac{1}{6}(4)+\frac{1}{6}(6)+\frac{1}{6}(8)+\frac{1}{6}(10)+\frac{1}{6}(-12)\\ &=\frac{2 + 4+6 + 8+10-12}{6}\\ &=\frac{(2 + 4+6 + 8+10)-12}{6}\\ &=\frac{30 - 12}{6}\\ &=\frac{18}{6}\\ &= 3 \end{align*}$$

\]

But wait, the cost to play should be equal to the expected value for the game to be fair. Wait, but let's check the options. Wait, maybe I made a mistake. Wait, the expected value of the winnings (without cost) should be equal to the cost. Let's recalculate:

Wait, when you roll 1: win 2, probability 1/6.

Roll 2: win 4, probability 1/6.

Roll 3: win 6, probability 1/6.

Roll 4: win 8, probability 1/6.

Roll 5: win 10, probability 1/6.

Roll 6: win - 12 (lose 12), probability 1/6.

So expected value \(E=\frac{2 + 4+6 + 8+10-12}{6}=\frac{30 - 12}{6}=\frac{18}{6}=3\). Wait, but the options: 3 tokens for one game. Wait, but let's check the other options. Wait, maybe I messed up. Wait, the expected value is the average gain, so for the game to be fair, the cost should be equal to the expected value. So if the expected value is 3 tokens (gain), then the cost to play should be 3 tokens per game? Wait, but let's check the options. The third option is "3 tokens for one game". But wait, let's check the other options. Wait, maybe I made a mistake in the expected value. Wait, no:

Sum of winnings for 1 - 5: 2 + 4 + 6 + 8 + 10 = 30. Winnings for 6: - 12. Total sum: 30 - 12 = 18. Divide by 6 (number of outcomes): 18/6 = 3. So the expected value is 3 tokens. So the cost to play should be 3 tokens per game for the game to be fair. So the correct option is "3 tokens for one game". But wait, let's check the options again. The options are:

1. 7 tokens for one game

1. 9 tokens for two games (which is 4.5 per game)

1. 3 tokens for one game

1. 1 token for two games (0.5 per game)

So the expected value is 3 per game, so the cost should be 3 per game. So the correct option is "3 tokens for one game".

Wait, but maybe I made a mistake. Let's re - express the expected value. The expected value of the net gain (winnings - cost) should be zero for a fair game. Let \(C\) be the cost. Then:

\(E(net)=E(winnings)-C = 0\)

We found \(E(winnings)=3\), so \(C = 3\). So the cost per game should be 3 tokens. So the correct option is "3 tokens for one game".

Answer:

3 tokens for one game